Timeline for Rationalizing and minimizing elliptic curve coefficients

6 events

when toggle format	what		by	license	comment
Dec 25, 2021 at 13:38	vote	accept	Maksym Voznyy
Dec 25, 2021 at 13:08	comment	added	Maarten Derickx		I.e. you could take $a=33z - 422$ where $z$ is a root of $x^2 - x + 6$.
Dec 25, 2021 at 13:02	comment	added	Maarten Derickx		@ChrisWuthrich Well $P_1$ is of norm 3, and $K$ contains a non principal element of norm 2 as well that is equivalent to $P_1$ in the class group. So you could reduce even further by taking $a=-33/2\sqrt{-23} - 811/2$. The only reason it doesn't look nicer is cause $\sqrt{-23}$ just isn't the nicest generator of $\mathbb Q(\sqrt{-23})$.
Dec 25, 2021 at 12:30	comment	added	Chris Wuthrich		$P_1$ is not principal either, so I don't see how you could get an integral $a$ of smaller norm. And that would be the best measure of having a "small" $a$ to me.
Dec 25, 2021 at 12:22	comment	added	Chris Wuthrich		On 2. The only transformation of a Weierstrass equation that leaves $a_1=a_3=a_6=0$ are of the form $x\to f\,x+r$ with $r$ either 0 or a solution to $r^2+ar+b=0$. Since $a^2-4b$ generates an ideal that is not a square in $K$, there are only those which divide $a$ by $f^2$.
Dec 25, 2021 at 10:57	history	answered	David Loeffler	CC BY-SA 4.0